Two Questions about Navigating as a Wave

Q1) What is the locus of points equidistant from two circles?

The distance from a point to a circle is understood to mean the shortest distance, which is along a straight line perpendicular to the arc.  On a circle, such a line passes through the origin of the circle... and the other side.  From any point of reference, we are interested in the nearest of these two intersections.

In general, given two sound sources propagating in air, at different locations, for an observer moving in relation to both sources, paths which are solutions to the locus problem are equal time curves.

Geometric solution:

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Two circles with centers A and B, radii r and s. The two loci are shown in blue and orange.  Click "Auxiliary Objects" to see the method of construction.

Identifying the cases is tricky with geometry.  Consider a partial construction:
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com
This appears to give the locus of all points F forming isosceles triangles DEF.  There are several problems.  For example, C lies on Ray AD and is undefined for s > r.  But if we use the Line through AD, it intersects α twice and  only one of these intersections at a time can be closest to us.  If F moves along line AD  it gives a second branch to the hyperbola, corresponding to D rotated beyond the point that the two rays AD, BE are parallel.  There is no such isosceles triangle with internal angles > 90°.
The solution is to let C move freely along the line AD, and restrict F to the ray.  Several problems remain, and then there is the question of the second intersection of Ray AD with β'.  Proceeding this way is klunky.  I'll use algebra.

Algebraic solution:

Two circles, α and β lie together on a sheet of paper.  They're just friends.  Their centers are the points A and B.  Draw coordinate axes on the table.  Connect the segment AB, bisect it and call the midpoint m.  Rotate the sheet until AB lies on the x-axis, then move it left or right until m lies on top of the coordinate origin.  O (0,0) bisects AB.  The coordinates of A and B are now A: (-AB/2, 0), (AB/2, 0).  For convenience, let the distance AB/2 = c.  The problem can now be written like this:

There are two circles  α and β with respective origins (-c, 0), (c, 0); and radii  r, s.   Solutions to the first equation form a circle t meters away from α.  Solutions to the second equation are t meters away from  β.  The values of t which satisfy both are the solution to the problem.  If the circles are advancing wave fronts, positive values of t are future times, and negative values are past.

There are four arrangements of the ± signs.

ORANGE:     a){(r + t), (s + t)}      b) {(r − t), (s − t)}
The solution in x and y is the same.  If t = k satisfies the first pair, t = −k satisfies the second.  Beginning with a), rewrite the first equation as y² = (r+t)² − (x+c)², and plug this value of y² into the second equation. Expanding the squares and simplifying, we have   .  Let

Then

Substitute this value of t back into (1), and again expanding the squares and simplifying, we obtain


In  the same manner, beginning with b), we obtain

with the same solution (4).

BLUE: c) {(r + t), (s − t)}       d) {(r − t), (s + t)}
The procedure is the same.  The results are

Which share the solution


Note that (3), (3b) and (4)  can be turned into (3c), (3d) and (4)' by trading the places of R and S.

Preliminary Result:  there are two distinct quadratic solutions of the same form, one determined by the difference of the two radii (R), the other by the sum (S):


Every line through the center of a circle determines two intersections: 1, 2.  Invariably, the closest, or principal, intersection is given by the vector from the center of the circle to our location.  The second intersection is the greatest distance from our location to the circle, and invariably crosses the origin of the circle.  Solutions to (4) and (4') are equidistant from a pair of these intersections... but which pair? There are four possibilities: (α₁,β₁), (α₂,β₂); and (α₁,β₂), (α₂,β₁).  We want to keep only the first .

ORANGE:
If  R² < 4c²,  (4) is a hyperbola.
   The branch closer to the origin of the smaller circle: (α₁,β₁), SOLUTION
   The second branch: (α₂,β₂).
If R² > 4c², (4) is an ellipse.  the equal lengths: (α₁,β₂), (α₂,β₁)
   according as α or β is greater.

BLUE:
If S² > 4c²,  (4)' is an ellipse.
    Equal distances: (α₁,β₁), SOLUTION
If S² < 4c², (4)' is a hyperbola.
    Equal distances of the two branches:  (α₁,β₂), (α₂,β₁)

In summary:

• ORANGE:   (4) is a solution only when  |R| < 2c
• BLUE:  (4)' is a solution only when S > 2c
• Cases:
• If one circle encloses the other, the solution is an ellipse (4)'.
• If the circles lie entirely outside one another, the solution is the branch of the hyperbola (4) closest to the smaller circle.
• If the circles intersect in two points, there are two solutions, one ellipse (4) and one branch of a hyperbola (4)'.
• The curvature of the locus changes continuously.  As r and s aproach equality, the curvature of the hyperbola goes to zero.  At r=s it is a line.
  Likewise, the curvature of the ellipse approaches a constant as the centers of the two circles approach.  When the centers of the circles coincide, the ellipse is a circle.
  These are not special cases, and part of providing a complete solution is accounting for them as continuous relationships (curvature).
• The three conditions sign(|R|− 2c); sign(S− 2c); sign(r − s) are sufficient to select the solutions we want.